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indefinite integral

paulmdrdo

Active member
May 13, 2013
386
how would i go about solving this

$\displaystyle\int\frac{\arctan x}{1+x^2}dx$?

i tried substitution but i didn't work.
 
Last edited:

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
That's very strange. Did you try the obvious substitution [tex]u= tan^{-1}(x)[/tex]. What is the derivative of [tex]tan^{1}(x)[/tex]?


(And do you understand that [tex]tan^{-1}(x)[/tex] here is the arctangent, NOT the [tex]\frac{1}{tan(x)}[/tex]?!)
 

paulmdrdo

Active member
May 13, 2013
386
yes, i know that. but how do i use substitution here?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
Following HoI's suggestion, we get that $u=\tan^{-1}(x)$, and
$$du= \frac{1}{1+x^{2}} \, dx,$$
and the integral becomes
$$\int u \, du.$$
Can you continue?
 

paulmdrdo

Active member
May 13, 2013
386
this is my answer,

$\displaystyle\frac{1}{2}(\tan^{-1}x)^2+C$
 
Last edited:

Plato

Well-known member
MHB Math Helper
Jan 27, 2012
196
this is my answer,

$\displaystyle\frac{1}{2}(Tan^{-1})^2+C$
That is the correct idea, but awful notation: $\displaystyle\frac{1}{2}(\arctan^{2}(x))+C$
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,143
That is the correct idea, but awful notation: $\displaystyle\frac{1}{2}(\arctan^{2}(x))+C$
I find that "atn(x)" works quite well also.

-Dan